A proof I was reading used the claim that $\vert\vert{N}\vert\vert_2$ = 1 for a nilpotent matrix $N$. I tried to prove it, and have a couple of questions on it.
First, my "proof": We know that there exists a basis wrt which the matrix of $N$ has $1$'s on the super-diagonal and $0$'s everywhere else. With this representation, for $x = (x_1, x_2, \cdots, x_n)^T$, $Nx = (0,x_2,x_3, \cdots, x_n)^T$
From the definition of operator norm, $||N||_p = sup||Nx||_p $ taken over $ ||x||_p=1$. Thus $||Nx||_p = 1 - |x_1|^p \leq 1$
Hence, $||N||_p \leq 1$. By choosing $x = (0, x_2, \cdots, x_n)^T$ we see that $||N||_p =1 $.
EDIT 1: An explicit choice $x = (0, 1,0, \cdots, 0)^T$ also works.
My questions are
- Is my proof correct?
- The norm seems to depend on the particular matrix representation used. Isn't this a problem? Can I say in general that "the operator norm of a nilpotent matrix is unity"?
EDIT 2: I read the book again and I see I misunderstood. The remark comes in the context that a Jordan block $J_i = \lambda_iI + N_i$, where $N_i$ is nilpotent and $||N_i||_2 = 1$. Of course, $N_i$ has the exact structure I used above in my proof, and it is never suggested that all nilpotent matrices have unit norm.